The radius of a cone is increasing at a rate of $3$ centimeters per second and the height of the cone is decreasing at a rate of $4$ centimeters per second. At a certain instant, the radius is $8$ centimeters and the height is $10$ centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{736\pi}{3}$ (Choice B) B $-\dfrac{224\pi}{3}$ (Choice C) C $\dfrac{224\pi}{3}$ (Choice D) D $\dfrac{736\pi}{3}$ The volume of a cone with radius $r$ and height $h$ is $\pi r^2\dfrac{h}{3}$.
Explanation: Setting up the math Let... $r(t)$ denote the radius at time $t$, $h(t)$ denote the height at time $t$, and $V(t)$ denote the volume at time $t$. We are given that $r'(t)=3$ and $h'(t)=-4$ (notice that $h'$ is negative). We are also given that that $r(t_0)=8$ and $h(t_0)=10$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures The measures relate to each other through the formula for the volume of a cone: $V(t)=\dfrac{\pi}{3}[r(t)]^2h(t)$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=\dfrac{\pi}{3}(2r(t)r'(t)h(t)+[r(t)]^2h'(t))$ Using the information to solve Let's plug ${r(t_0)}={8}$, ${r'(t_0)}={3}$, ${h(t_0)}={10}$, and $C{h'(t_0)}=C{-4}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=\dfrac{\pi}{3}(2{r(t_0)}{r'(t_0)}{h(t_0)}+[{r(t_0)}]^2C{h'(t_0)}) \\\\ &=\dfrac{\pi}{3}(2({8})({3})({10})+({8})^2(C{-4})) \\\\ &=\dfrac{224\pi}{3} \end{aligned}$ In conclusion, the rate of change of the volume of the cone at that instant is $\dfrac{224\pi}{3}$ cubic centimeters per second. Since the rate of change is positive, we know that the volume is increasing.